Question

for each problem, find the: x and y intercepts, x - coordinates of the critical points, open intervals where the function is increasing and decreasing, x - coordinates of the inflection points, open intervals where the function is concave up and concave down, and relative minima and maxima. using this information, sketch the graph of the function.

1. ( y = -\frac{x^{3}}{3}+x^{2} )

Explanation:

Step1: Find \(x\) and \(y\) - intercepts

• \(y\)-intercept:

Set \(x = 0\). Then \(y=-\frac{0^{3}}{3}+0^{2}=0\).

• \(x\)-intercept:

Set \(y = 0\), so \(0=-\frac{x^{3}}{3}+x^{2}\). Factor out \(x^{2}\): \(x^{2}(1 - \frac{x}{3})=0\).
Solving \(x^{2}=0\) gives \(x = 0\), and solving \(1-\frac{x}{3}=0\) gives \(x = 3\).

Step2: Find the first - derivative and critical points

The function \(y=-\frac{x^{3}}{3}+x^{2}\).
The first - derivative \(y^{\prime}=-x^{2}+2x\).
Set \(y^{\prime}=0\): \(-x^{2}+2x = 0\). Factor out \(-x\): \(-x(x - 2)=0\).
The critical points are \(x = 0\) and \(x = 2\).

Step3: Determine intervals of increase and decrease

Use the test - point method.
The intervals are \((-\infty,0)\), \((0,2)\), and \((2,\infty)\).

• For \(x=-1\) (in \((-\infty,0)\)): \(y^{\prime}=-(-1)^{2}+2(-1)=-1 - 2=-3<0\).
• For \(x = 1\) (in \((0,2)\)): \(y^{\prime}=-1^{2}+2\times1=1>0\).
• For \(x = 3\) (in \((2,\infty)\)): \(y^{\prime}=-3^{2}+2\times3=-9 + 6=-3<0\).

The function is decreasing on \((-\infty,0)\cup(2,\infty)\) and increasing on \((0,2)\).

Step4: Find the second - derivative and inflection points

The second - derivative \(y^{\prime\prime}=-2x + 2\).
Set \(y^{\prime\prime}=0\): \(-2x+2 = 0\), then \(x = 1\).

Step5: Determine concavity

• For \(x<1\) (e.g., \(x = 0\)): \(y^{\prime\prime}=-2\times0+2=2>0\).
• For \(x>1\) (e.g., \(x = 2\)): \(y^{\prime\prime}=-2\times2+2=-2<0\).

The function is concave up on \((-\infty,1)\) and concave down on \((1,\infty)\).

Step6: Find relative minima and maxima

Use the first - derivative test.

• At \(x = 0\): Since the function changes from decreasing (\(x<0\)) to increasing (\(0

Since \(y^{\prime}\) changes sign from negative (\(x<0\)) to positive (\(0Since \(y^{\prime}\) changes sign from positive (\(02\)), \(y(2)=-\frac{2^{3}}{3}+2^{2}=-\frac{8}{3}+4=\frac{4}{3}\) is a relative maximum.

Answer:

• \(x\)-intercepts: \(x = 0\) and \(x = 3\); \(y\)-intercept: \(y = 0\).
• Critical points: \(x = 0\) and \(x = 2\).
• Intervals of increase: \((0,2)\); Intervals of decrease: \((-\infty,0)\cup(2,\infty)\).
• Inflection point: \(x = 1\).
• Concave up: \((-\infty,1)\); Concave down: \((1,\infty)\).
• Relative minimum: At \(x = 0\), \(y = 0\); Relative maximum: At \(x = 2\), \(y=\frac{4}{3}\).

To sketch the graph:

• Plot the intercepts \((0,0)\) and \((3,0)\).
• Mark the critical points \((0,0)\) (relative minimum) and \((2,\frac{4}{3})\) (relative maximum).
• Mark the inflection point \((1,y(1))\) where \(y(1)=-\frac{1}{3}+1=\frac{2}{3}\).
• Use the information about concavity (concave up on \((-\infty,1)\) and concave down on \((1,\infty)\)) and intervals of increase/decrease to draw a smooth curve.